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\title{MATH0037 Galois Theory 2002, Sheet 8}
\author{Geoff Smith, http://www.bath.ac.uk/$\sim$masgcs}
\date{}
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\maketitle
\begin{enumerate}
\item {\em Let $\alpha = \sqrt[3] 2$ and $\omega = e^{2\pi i/3}$.
Let $K = \mathbb Q[\alpha],\ L = \mathbb Q[\omega]$
and $M = K[\omega] = L[\alpha]$.
Note that $M$ is the smallest subfield of $\mathbb C$ which contains
both $\omega$ and $\alpha$.}
\begin{enumerate}
\item[(a)] {\em Prove that $|K:\mathbb Q| = 3$ and $|L: \mathbb Q| = 2$.} \newline
\noindent {\bf Solution}\ $\alpha$ is a root of $X^3 -2$ which is irreducible
as an element of $\mathbb Q[X]$ by Eisenstein's criterion. This shows that
dim$_{\mathbb Q} K = 3.$ Similarly $\omega$ is a root of $X^2 + X + 1$
which is also irreducible in $\mathbb Q[X]$ else $\omega$ would be real
(which it is not).
\item[(b)] {\em Observe that $\omega \not \in \mathbb R$ so $\omega \not \in
K$. Deduce that $|M:K| = 2$ and so $|M:\mathbb Q| = 6$.} Now $\omega$ is a root
of $X^2 + X + 1$ which is irreducible as an element of $K[X]$
else $\omega$ would be in $K$ and therefore would be real which it is not.
Note that $M = K(\omega)$ so $|M:K|=2$ and 
$|M:\mathbb Q| = |M:K| \cdot |K:\mathbb Q| = 2 \times 3 = 6.$
\item[(c)] Suppose that $\beta  \in \mathbb C$ is any root of
$X^3 -2$. Show that $L[\beta] = M$.\newline
\noindent {\bf Solution }\ Now $\beta$ is a root of $X^3 -2$ which
is irreducible as an element of $\mathbb Q[X]$ by Eisenstein's
criterion, so $|\mathbb Q(\beta): \mathbb Q| = 3$.
\item {\em Suppose that $\beta  \in \mathbb C$ is any root of
$X^3 -2$. There is a natural isomorphism $L[X]/(X^3-2) \rightarrow L[\beta]$.
Let $\beta_1, \beta_2$ be two distinct roots of
$X^3 -2$ in $\mathbb C$. Deduce that there is an element $g$ in 
Gal$_L M$ such that $(\beta_1) g = \beta_2$. Show that $g$ has order 3.}
\newline 
\noindent {\bf Solution }\ Each possible $\beta$ is an element of $M$
and since $|L(\beta):L| =3 $ and $L \leq L(\beta) \leq M$ and $|M:L| = 3$,
it follows that $M = L(\beta).$ Now
$M = L[\beta_1] \simeq L[X]/(X^3-2) \simeq L[\beta_2] = M$
and composing the isomorphisms gives us an element $g \in \mbox{Gal}_L M$
such that $(\beta_1)g = \beta_2$. Since $\beta_1 \not = \beta_2$ it follows
that $g$ is not the identity element of this group. Moreover $g$ fixes $L$ elementwise
so $(\omega)g = \omega$. The roots of $X^3-2$ are $\alpha, \omega \alpha$ and   
$\omega^2 \alpha$. The ratio of any pair of these is $\omega$ or $\omega^{-1}$.
Replacing $g$ by $g^{-1}$ if necessary, we may assume that $\beta_2 = \omega \beta_1$.
Therefore \[(\beta_2)g = (\beta_1 \omega)g = \beta_2 \omega = \beta_1\omega^2 \not
= \beta_1.\]
Finally \[ (\beta_1\omega^2)g = (\beta_1 \omega)\omega^2 = \beta_1.\]
Now $M = L(\beta_1)$ so $g^3 =1$. This is the smallest power of $g$
which is the identity, so the order of $g$ is 3.
\item[(d)] {\em Show that complex conjugation induces an element $c$ in 
Gal$_K M$ of order $2$.} \newline
\noindent {\bf Solution} Certainly complex conjugation swaps
$\omega$ and $\omega^2$ and fixes $K$ elementwise. Thus this induces
an automorphism of $M$ which is not the identity but with square
which is the identity.
\item[(e)] {\em Deduce that Gal$_{\mathbb Q} M$ contains an element
of order 2 and an element of order 3.} The automorphism $g$ constructed 
in part (c) belongs to Gal$_{\mathbb Q}M$, and combined with the 
conclusion of part (d), we are done. 
\item[(f)] {\em Notice that $\Omega = \{ \alpha, \omega \alpha, \omega^2 \alpha\}$
is the set of complex roots of $X^3 -2$. Let $S_\Omega$ denote the
group of bijections from $\Omega$ to $\Omega$ (under composition 
of maps). Suppose that 
$h \in $ Gal$_{\mathbb Q} M$. Show that $(x)h \in \Omega$ for 
all $x \in \Omega$, and deduce that the application of $h$
induces an element $\widehat h \in S_\Omega$. Show that
the map $\eta:$ Gal$_{\mathbb Q} M \rightarrow S_\Omega$
defined by $h \mapsto \widehat h$ is a group homomorphism.}\newline
\noindent {\bf Solution} We have covered this in lectures.
\item[(g)] {\em Show that the homomorphism $\eta$ has trivial kernel,
so that $\eta$ is a monomorphism.} If an element of Gal$_{\mathbb Q}M$ fixes
the roots of $X^3 -2$, then it fixes both $\alpha$ and $\omega \alpha$,
and therefore fixes $\omega$. Now $M = \mathbb Q(\omega, \alpha)$ so
$g$ fixes each element of $M$. 
\item[(h)] {\em Use the fact that Gal$_{\mathbb Q} M$ contains elements of
order 2 and 3 to decide that $\eta$ is surjective, and is therefore
an isomorphism.}\newline
\noindent {\bf Solution}\ This injective homomorphism has image 
which contains elements of order both 2 and 3 in a group of order 6.
By Lagrange's theorem, 2 and 3 both divide the order of the Galois group, and this
order must divide 6. We deduce that $\eta$ is bijective and Gal$_{\mathbb Q}M$
is a group of order 6 (and a copy of $S_3$).
\end{enumerate}

\item {\em Let $K$ denote the smallest subfield of $\mathbb C$ which contains
both $\sqrt 2$ and $\sqrt 3$. Show that $|K : \mathbb Q| = 4$.
By echoing the method of the previous paragraph, show that
Gal$_{\mathbb Q} K$ is a group of size 4.} {\rm (It is recommended
to use a new $\Omega = \{ \sqrt 2, -\sqrt 2, \sqrt 3, -\sqrt 3\}$
to replace the old $\Omega$.)}\newline \noindent
{\bf Solution}\ Let $\delta = \sqrt 2 + \sqrt 3$ so $\delta^2 -5 = 2 \sqrt 6$.
Thus $(\delta^2 -5)^2 = 24$ and so $\delta$ is a root of $f = X^4 -10X^2 +1$. 
Since the four roots are $\sqrt 2 + \sqrt 3,\ \sqrt 2 - \sqrt 3$ and 
$-\sqrt 2 - \sqrt 3$, none of which is rational, any possible factorization
of our polynomial in $\mathbb Q[X]$ must be into two quadratic, and the
only possibility of a rational factorization into monic polynomials
will be as $(X - \sqrt 2 - \sqrt 3)(X + \sqrt 2 + \sqrt 3)$ by
$(X - \sqrt 2 + \sqrt 3)(X + \sqrt 2 -\sqrt 3)$, but expansion reveals
that neither of these quadratics has rational coefficients. Thus
$f$ is irreducible as a rational polynomial. It follows that the maps
$\sqrt 2 + \sqrt 3 \mapsto \epsilon$ as $\epsilon$ runs through the
roots of $f$ yield the four elements of the Galois group
Gal$_{\mathbb Q}K$. The action of $G$ on $\Omega$ reveals that the four
elements of $G$ act on $\Omega$ as follows: the identity;
swap $\sqrt 2$ and $-\sqrt 2$ but fix $\sqrt 3$; swap
$\sqrt 3$ and $-\sqrt 3$ but fix $\sqrt 2$; finally swap $\sqrt 2$ with $-\sqrt 2$
and simultaneously swap $\sqrt 3$ with $-\sqrt 3$.
\end{enumerate}

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